Algebra 1 · A.APR.B.3 · A.SSE.B.3A

Roots, zeros and x-intercepts

An equation has solutions, a function has zeros and a graph has x-intercepts. Why do all three turn out to be the same numbers?

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Graphing from standard formAlgebra 1

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Quadratic inequalitiesAlgebra 2 Intercepts and zeros of a functionAlgebra 1 Factored formAlgebra 1 The discriminant and the number of real solutionsAlgebra 1 +1 more

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Roots, zeros and x-intercepts

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Why this exists

The same numbers keep turning up under different names. Solve x^2 - 2x - 3 = 0 and you get -1 and 3, the roots of the equation. Ask which inputs make f(x) = x^2 - 2x - 3 output zero and you get -1 and 3 again, the zeros of the function. Draw the graph and it meets the x-axis at -1 and 3, the x-intercepts.

These are not three facts that happen to agree. They are one fact seen three ways: in the equation, in the function, and in the picture. Once you see why, you can move freely between the three. You can solve by factoring to find where a graph crosses the axis, read a rough answer off a graph before doing any algebra, and check a solution by looking at a picture.

Definition

Take a function f and a real number k.

k is a root of the equation f(x) = 0 when substituting k for x makes the equation true. “Root” and “solution” mean the same thing here; “root” is the older word, and it is the usual one for polynomial equations.
k is a zero of the function f when f(k) = 0: the input k gives the output 0.
The point (k, 0) is an x-intercept of the graph y = f(x) when the graph passes through it: the graph meets the x-axis there.

Roots and zeros are numbers. An x-intercept is a point, and its y-coordinate is always 0. Many books also call the number k itself the x-intercept; the meaning is clear either way, but the point is the exact statement.

The y-intercept is a different thing: the point (0, f(0)) where the graph meets the y-axis, found by putting 0 in, not by getting 0 out. See the y-intercept.

Intuition

A graph is a way of storing a function as a picture. For each input x it plots the point (x, f(x)), so the output is drawn as a height above or below the x-axis. In Figure 1, the input 4 gives the output f(4) = 16 - 8 - 3 = 5, so the graph is 5 units up at x = 4.

Figure 1The height of the graph is the output, and on the x-axis that height is zero.

The x-axis is the line of height zero. So the question “where does the graph meet the x-axis?” is the question “for which inputs is the height zero?”, which is the question “for which x is f(x) = 0?”. That last question is an equation, and its answers are its roots.

The equation is the algebra, the function is the input and output, the graph is the picture, and all three are asking the same thing.

Results

The three names agree. For a function f and a real number k, these three statements say the same thing:

k is a root of the equation f(x) = 0;
f(k) = 0, that is, k is a zero of f;
(k, 0) is an x-intercept of the graph y = f(x).

Zeros and factors. For a quadratic f(x) = ax^2 + bx + c and a real number r, r is a zero of f exactly when (x - r) is a factor of f(x). So in factored form

f(x) = a(x - r)(x - s)

the zeros are r and s, and nothing else.

At most two. A quadratic has at most two zeros, so its graph meets the x-axis at most twice.

The axis sits halfway. If a quadratic has zeros r and s, its axis of symmetry is the vertical line halfway between them:

x = \frac{r + s}{2}

Why it is true

The three names. The graph of f is the set of points (x, y) with y = f(x). A point lies on the x-axis exactly when its y-coordinate is 0. So (k, 0) is on the graph exactly when 0 = f(k), which is the definition of a zero, and “f(k) = 0” is exactly what it means for k to make the equation f(x) = 0 true.

A factor gives a zero. If f(x) = a(x - r)(x - s), then f(r) = a \cdot 0 \cdot (r - s) = 0. Going the other way, by the zero product property, f(x) = 0 forces x - r = 0 or x - s = 0, since a \neq 0. So the zeros are r and s and no other numbers.

A zero gives a factor. Suppose f(r) = 0. Subtracting zero changes nothing, so

\begin{aligned} f(x) &= f(x) - f(r) \\ &= (ax^2 + bx + c) - (ar^2 + br + c) \\ &= a(x^2 - r^2) + b(x - r) \\ &= (x - r)\big(a(x + r) + b\big) \end{aligned}

The c cancels, and both remaining pieces contain (x - r), which is why it factors out.

At most two. Once (x - r) is factored out, the other factor a(x + r) + b is linear with a \neq 0, so it is zero for exactly one x. By the zero product property, f is zero at r and at that one other number, which may be r again, and nowhere else.

Halfway. With f(x) = a(x - r)(x - s), put r + s - x in place of x:

\begin{aligned} f(r + s - x) &= a(s - x)(r - x) \\ &= a(x - r)(x - s) = f(x) \end{aligned}

So the inputs x and r + s - x always give the same output. Those two inputs are mirror images across \displaystyle x = \frac{r + s}{2}, their average, so the graph is symmetric about that line.

Worked examples

Example 1

Reading zeros from factored form

Find the zeros of f(x) = (x - 2)(x + 5), the x-intercepts of its graph, and its y-intercept.

Solution

A zero is an input that makes the output 0, so set the product equal to 0:

(x - 2)(x + 5) = 0

A product is zero only when a factor is zero, so x - 2 = 0 or x + 5 = 0, giving x = 2 or x = -5. The zeros are 2 and -5, the roots of the equation are 2 and -5, and the x-intercepts are (2, 0) and (-5, 0).

The sign flips: the factor x + 5 gives the zero -5, because -5 is the number that makes x + 5 zero.

The y-intercept comes from putting 0 in: f(0) = (0 - 2)(0 + 5) = -10, so it is (0, -10). It is not a zero, and it is found the other way round.

Example 2

Factoring first, then sketching

Find the zeros of f(x) = x^2 - 2x - 8, and use them to find the vertex and sketch the graph.

Solution

The zeros are the roots of x^2 - 2x - 8 = 0. We need two numbers whose product is -8 and whose sum is -2; they are -4 and 2:

x^2 - 2x - 8 = (x - 4)(x + 2)

So the zeros are 4 and -2, and the x-intercepts are (4, 0) and (-2, 0).

The axis of symmetry lies halfway between the zeros:

x = \frac{-2 + 4}{2} = 1

The vertex is on the axis, so its x-coordinate is 1 and its height is f(1) = 1 - 2 - 8 = -9. The vertex is (1, -9). The y-intercept is f(0) = -8, the point (0, -8).

Figure 2The two zeros fix the axis of symmetry halfway between them, and the vertex sits on it.

Four points and a line of symmetry are enough for an honest sketch. This is the shortcut that graphing from standard form uses whenever the quadratic factors.

Example 3

A leading coefficient that is not one

Find the zeros of f(x) = 2x^2 + 5x - 3.

Solution

Factor. The first terms must multiply to 2x^2 and the last to -3, and the cross terms must add to 5x:

2x^2 + 5x - 3 = (2x - 1)(x + 3)

Check by expanding: 2x^2 + 6x - x - 3 = 2x^2 + 5x - 3.

Now set each factor to zero. 2x - 1 = 0 gives \displaystyle x = \frac{1}{2}, and x + 3 = 0 gives x = -3. The zeros are \displaystyle \frac{1}{2} and -3.

The zero of 2x - 1 is \displaystyle \frac{1}{2}, not 1: it is whatever makes the factor zero, and that takes solving. A check confirms it: \displaystyle f\left(\frac{1}{2}\right) = \frac{2}{4} + \frac{5}{2} - 3 = 0.

Example 4

Two zeros, one, or none

Find the real zeros of f(x) = x^2 - 6x + 5, g(x) = x^2 - 6x + 9 and h(x) = x^2 - 6x + 13.

Solution

f(x) = (x - 1)(x - 5), so f has two zeros, 1 and 5.

g(x) = (x - 3)^2. Both factors are the same, so g has one zero, 3. Its graph touches the axis at (3, 0) and turns back instead of crossing.

h does not factor with integers: no two integers multiply to 13 and add to -6. Split off the perfect square instead:

h(x) = (x^2 - 6x + 9) + 4 = (x - 3)^2 + 4

A square is never negative, so h(x) \geq 4 for every real x. The output is never 0, so h has no real zeros and its graph never meets the x-axis.

The equation h(x) = 0 still has roots, just not real ones. (x - 3)^2 = -4 gives x - 3 = \pm 2i, so x = 3 \pm 2i, which are complex numbers. They are roots, but there is no point for them on the real graph.

Figure 3The graphs of f, g and h from Example 4. Lifting a parabola changes how many zeros it has, from two, to one, to none that are real.

The three graphs are the same parabola lifted 4 units at a time. The vertex rises through the axis, and the number of x-intercepts drops from two to one to none.

Example 5

Building a quadratic from its zeros

Find the quadratic function whose zeros are -1 and 4 and whose graph meets the y-axis at (0, 8).

Solution

Each zero r gives a factor (x - r), so the function must be

f(x) = a(x + 1)(x - 4)

for some number a \neq 0. Any value of a keeps the same zeros, because multiplying 0 by a still gives 0; what a changes is how steep the parabola is and which way it opens.

The y-intercept fixes a. Put in x = 0:

f(0) = a(1)(-4) = -4a = 8 \qquad\Longrightarrow\qquad a = -2

So f(x) = -2(x + 1)(x - 4). Multiplied out, f(x) = -2(x^2 - 3x - 4) = -2x^2 + 6x + 8.

Check. f(0) = 8, f(-1) = -2 - 6 + 8 = 0 and f(4) = -32 + 24 + 8 = 0. The negative a means the parabola opens downward.

Example 6

Where a line meets a parabola

Find the points where the parabola y = x^2 - 2x - 2 meets the line y = x + 2.

Solution

A point on both graphs has the same y for the same x, so the two outputs are equal there:

x^2 - 2x - 2 = x + 2

Move everything to one side, so that the question becomes one about zeros:

\begin{aligned} x^2 - 3x - 4 &= 0 \\ (x - 4)(x + 1) &= 0 \end{aligned}

So x = 4 or x = -1. The line gives the heights: 4 + 2 = 6 and -1 + 2 = 1. The graphs meet at (4, 6) and (-1, 1).

Check. On the parabola, 16 - 8 - 2 = 6 and 1 + 2 - 2 = 1. Both agree.

Figure 4The graphs meet directly above -1 and 4, the zeros of x^2 - 3x - 4.

The numbers 4 and -1 are the zeros of the gap d(x) = (x^2 - 2x - 2) - (x + 2), the height of the parabola above the line. Where the gap is zero, the graphs meet. Every equation f(x) = g(x) is a question about the zeros of f(x) - g(x), which is why moving everything to one side is always the first step.

Boundaries

Only real zeros appear on the graph. A quadratic can have roots that are complex numbers, as h did in Example 4. They solve the equation, but the graph only shows real inputs, so they give no x-intercept.

The other side must be zero. The roots of x^2 - 2x = 8 are the zeros of x^2 - 2x - 8, not of x^2 - 2x. Zeros belong to the function you get once everything is on one side.

A graph gives estimates, not exact values. The zeros of x^2 - 2x - 2 are \displaystyle 1 \pm \sqrt{3}, about -0.73 and 2.73. A graph shows them near there; only algebra, or the quadratic formula, gives them exactly.

One intercept can count twice. g(x) = (x - 3)^2 has one x-intercept, but the factor (x - 3) appears twice. Whether a zero is counted once or twice, and how that shows as crossing or touching, is multiplicity.

The factor rule here is for polynomials. For other functions, zeros and x-intercepts still agree, since that follows from what a graph is, but “zero means factor” is a fact about polynomials, proved in general by the factor theorem.

Where this goes next

Written as a(x - r)(x - s), a quadratic shows its zeros at a glance, which is the point of factored form. Going the other way, from zeros to a function, is building a polynomial from its zeros.

Whether a quadratic has two zeros, one, or none can be told without solving, from a single number built from its coefficients; that is the discriminant and the graph. The zeros are also the only places a quadratic can change sign, which is the whole method of quadratic inequalities.

For polynomials of higher degree, the same three names still agree, and the zeros and how often each repeats shape the whole graph (zeros and multiplicity). And when a function cannot be factored at all, a change of sign on a continuous graph still proves a zero is there: showing a root exists.

Exercises

From factored form

Find the zeros of each function and the x-intercepts of its graph.

f(x) = (x - 6)(x + 1)
g(x) = (x + 7)(x + 2)
h(x) = x(x - 9)
p(x) = (3x - 2)(x + 4)
q(x) = -2(x - 5)^2

Factor first

Find the real zeros of each function.

f(x) = x^2 + x - 12
f(x) = x^2 - 10x + 21
f(x) = x^2 - 49
f(x) = 2x^2 - 9x + 4
f(x) = 3x^2 + 12x
f(x) = 4x^2 - 12x + 9
f(x) = x^2 + 2x + 5
For f(x) = x^2 - 4x - 5, find the zeros, the x-intercepts, the y-intercept and the vertex.

Graphs, equations and functions

Figure 5The graph of g for Exercise 14.

Figure 5 shows the graph of a quadratic function g. State the zeros of g, then find g(x) in standard form.
Write, in standard form, the quadratic with leading coefficient 1 whose zeros are 2 and 6.
Find the quadratic function whose zeros are -1 and 3 and whose graph passes through (1, 8).
Find the points where the parabola y = x^2 - 4x + 1 meets the line y = 2x - 4.
Find the zeros of f(x) = x^3 - 9x.
For f(x) = x^2 + 2x - 8, find every x for which f(x) = 7.

Reasoning

A quadratic has zeros -4 and 10. Explain why its vertex has x-coordinate 3.
The equation x^2 + 9 = 0 has two roots, but the graph of y = x^2 + 9 has no x-intercepts. Explain how both can be true.
Show that if 5 is a zero of f(x) = x^2 + bx + c, then (x - 5) is a factor of f(x).
Can a quadratic function have three different zeros? Explain.
Explain why the graph of a function has at most one y-intercept but can have many x-intercepts.

Answers

Zeros 6 and -1; x-intercepts (6, 0) and (-1, 0). x - 6 = 0 at x = 6 and x + 1 = 0 at x = -1.
Zeros -7 and -2; x-intercepts (-7, 0) and (-2, 0). Both factors are sums, so both zeros are negative.
Zeros 0 and 9; x-intercepts (0, 0) and (9, 0). The factor x is zero at 0, so the graph passes through the origin.
Zeros \displaystyle \frac{2}{3} and -4; x-intercepts \displaystyle \left(\frac{2}{3}, 0\right) and (-4, 0). 3x - 2 = 0 gives \displaystyle x = \frac{2}{3}.
One zero, 5; one x-intercept, (5, 0). The constant -2 is never zero, and the graph touches the axis at (5, 0) without crossing.
-4 and 3. x^2 + x - 12 = (x + 4)(x - 3).
3 and 7. x^2 - 10x + 21 = (x - 3)(x - 7).
7 and -7. x^2 - 49 = (x - 7)(x + 7).
\displaystyle \frac{1}{2} and 4. 2x^2 - 9x + 4 = (2x - 1)(x - 4).
0 and -4. 3x^2 + 12x = 3x(x + 4).
\displaystyle \frac{3}{2} only. 4x^2 - 12x + 9 = (2x - 3)^2, so the graph touches the axis once.
No real zeros. x^2 + 2x + 5 = (x + 1)^2 + 4 \geq 4, so the output is never 0. Over the complex numbers the roots are -1 \pm 2i.
Zeros 5 and -1; x-intercepts (5, 0) and (-1, 0); y-intercept (0, -5); vertex (2, -9). f(x) = (x - 5)(x + 1), the axis is \displaystyle x = \frac{5 + (-1)}{2} = 2, and f(2) = 4 - 8 - 5 = -9.
Zeros -3 and 1, and g(x) = -x^2 - 2x + 3. So g(x) = a(x + 3)(x - 1), and the y-intercept gives g(0) = -3a = 3, so a = -1.
x^2 - 8x + 12. It is (x - 2)(x - 6) multiplied out.
f(x) = -2x^2 + 4x + 6. f(x) = a(x + 1)(x - 3) and f(1) = a(2)(-2) = -4a = 8, so a = -2.
(1, -2) and (5, 6). x^2 - 4x + 1 = 2x - 4 gives x^2 - 6x + 5 = 0, so (x - 1)(x - 5) = 0; the line gives the heights 2(1) - 4 = -2 and 2(5) - 4 = 6.
-3, 0 and 3. x^3 - 9x = x(x^2 - 9) = x(x - 3)(x + 3).
x = -5 or x = 3. x^2 + 2x - 8 = 7 gives x^2 + 2x - 15 = 0, so (x + 5)(x - 3) = 0.
The axis of symmetry is halfway between the zeros, at \displaystyle x = \frac{-4 + 10}{2} = 3, and the vertex lies on the axis.
The roots are 3i and -3i, since (\pm 3i)^2 = -9. They are complex, not real, and a graph on the real plane shows only real inputs, so there is no point (k, 0) for them.
Since f(5) = 0, subtracting it changes nothing: f(x) = f(x) - f(5). The c cancels, leaving (x^2 - 25) + b(x - 5), and both parts contain x - 5, so f(x) = (x - 5)(x + 5 + b).
No. If r is a zero then f(x) = (x - r)(\text{a linear factor}), and a linear factor has only one zero, so there are at most two zeros in all.
A function gives exactly one output for the input 0, so the graph meets the y-axis at most once. Many different inputs can give the output 0, so the graph can meet the x-axis many times.